# Using the Schrodinger eqn in finding the momentum operator

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• Hamiltonian
In summary, the Schrodinger equation does not have a formal derivation and it is simply obtained by applying the Hamiltonian operator on the wave function. To derive the equation for momentum, we substitute the momentum operator in the Hamiltonian and use the Schrodinger equation to find its value. However, there are other ways to motivate the definition of momentum, such as considering it as the generator of spatial translations.
Hamiltonian
TL;DR Summary
how can we use the Schrodinger equation while finding ##\hat p## when in fact we have already used ##\hat p##(i.e. ##\hat p ^2## in the ##\hat T## term of the ##\hat H##) in the Schrodinger equation?
I have read that the Schrodinger equation has no formal derivation we are simply applying the Hamiltonian operator on the wave function
$$\hat H = i\hbar \frac{\partial}{\partial t} = \hat T + \hat V$$
here we substitute $$\hat T = \frac{\hat p^2}{2m}$$ where $$\hat p = -i \hbar \frac{\partial}{\partial x}$$
but when we derive the equation for ##\hat p## we actually substitute ##\frac{\partial \psi}{\partial t}## and ##\frac{\partial \psi *}{\partial t}## from the Schrodinger equation.

$$< p> = m\frac{d<x>}{dt} = m\int_{-\infty}^{+\infty} x\frac{\partial (\psi^*\psi)}{\partial t}$$
$$<p> = m\int_{-\infty}^{+\infty} x[\psi^*\frac{\partial \psi}{\partial t}+\psi\frac{\partial \psi^*}{\partial t}] dx$$
here we substitute ##\frac{\partial \psi}{\partial t}## and ##\frac{\partial \psi *}{\partial t}## as
$$\frac{\partial \psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^2 \psi}{\partial x^2} -\frac{i}{\hbar} V\psi$$
$$\frac{\partial \psi^*}{\partial t} = \frac{-i\hbar}{2m}\frac{\partial^2 \psi^*}{\partial x^2} +\frac{i}{\hbar} V\psi^*$$
after some simplification we end up with
$$<p> = \int_{-\infty}^{+\infty} \psi^* (-i\hbar \frac{\partial}{\partial x})\psi dx$$
and then finally we get $$\hat p = -i\hbar \frac{\partial }{\partial x}$$
so I don't understand how we can use the Schrodinger equation while finding ##\hat p## when in fact we have already used ##\hat p##(i.e. ##\hat p ^2## in the ##\hat T## term of the ##\hat H##) in the Schrodinger equation?
this video does the derivation for the momentum operator

Last edited:
What you appear to be showing is that$$m\frac{d\langle x \rangle}{dt} = \langle p \rangle$$ where ##\hat p = -i\hbar \frac{\partial}{\partial x}##. And that justifies the original definition of ##\hat p##.

vanhees71
PeroK said:
What you appear to be showing is that$$m\frac{d\langle x \rangle}{dt} = \langle p \rangle$$ where ##\hat p = -i\hbar \frac{\partial}{\partial x}##. And that justifies the original definition of ##\hat p##.
how exactly do we find ##\hat p## without using the Schrodinger equation? by finding ##\hat p## I mean how do we arrive at ##\hat p = -i\hbar \frac{\partial}{\partial x}##. I thought the only way of arriving at this would be by using ##m\frac{d<x>}{dt} = <p>## but when we use this approach we need to use the schrodinger equation but the KE energy term in the Hamiltonian is already using ##\hat p = -i\hbar \frac{\partial}{\partial x}##

vanhees71
Hamiltonian299792458 said:
how exactly do we find ##\hat p## without using the Schrodinger equation? by finding ##\hat p## I mean how do we arrive at ##\hat p = -i\hbar \frac{\partial}{\partial x}##. I thought the only way of arriving at this would be by using ##m\frac{d<x>}{dt} = <p>## but when we use this approach we need to use the schrodinger equation but the KE energy term in the Hamiltonian is already using ##\hat p = -i\hbar \frac{\partial}{\partial x}##
There are ways to motivate the definition of momentum. For example:

https://en.wikipedia.org/wiki/Momentum_operator#Origin_from_De_Broglie_plane_waves

And also on that page momentum as the generator of spatial translations (this is done in Sakurai's book).

vanhees71 and Hamiltonian

## 1. What is the Schrodinger equation?

The Schrodinger equation is a fundamental equation in quantum mechanics that describes the time evolution of a quantum system. It is used to determine the wave function, which contains all the information about the system's properties and behavior.

## 2. How is the Schrodinger equation used to find the momentum operator?

The Schrodinger equation can be used to find the momentum operator by taking the derivative of the wave function with respect to position. This operator represents the observable quantity of momentum in the quantum system.

## 3. What is the significance of the momentum operator in quantum mechanics?

The momentum operator is significant in quantum mechanics because it allows us to measure and predict the momentum of a quantum system. It is also a key component in the Heisenberg uncertainty principle, which states that the more precisely we know the position of a particle, the less precisely we know its momentum.

## 4. Are there any limitations to using the Schrodinger equation to find the momentum operator?

Yes, there are limitations to using the Schrodinger equation in finding the momentum operator. It can only be applied to systems that are in a stationary state, meaning that the system's properties do not change over time. Additionally, the Schrodinger equation does not account for relativistic effects, so it is not applicable to high-speed particles.

## 5. How is the momentum operator related to the Hamiltonian operator?

The momentum operator and the Hamiltonian operator are related through the commutator, which is a mathematical operation that represents how two operators behave when applied to a wave function. The commutator of the momentum operator and the Hamiltonian operator is equal to the negative of the position operator, which is another key operator in quantum mechanics.

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